Savvas Learning Co Geometry Student Edition Chapter 5 Relationships Within Triangles Exercise

Savvas Learning Co Geometry Student Edition Chapter 5 Relationships Within Triangles Exercise

Savvas Learning Co Geometry Student Edition Chapter 5 Relationships Within Triangles Exercise Solutions Page 281  Exercise 1  Problem 1

Given: Use a compass and straightedge

To find –  Construct the perpendicular bisector of a segment.

Open the compass more than half of the distance between A and B n, and scribe arcs of the same radius centered at A and B.

Call the two points where these two arcs meet C and D. Draw the line between C and D.

CD is the perpendicular bisector of the line segment AB.

We have

The perpendicular bisector of a segment:

Chapter 5 Relationships Within Triangles Exercise Savvas Geometry Answers Page 281  Exercise 2  Problem 2

Given: A(5,1),B(−3,3),C(1,−7).

To find –  Find the coordinates of the midpoints of the sides and then find the lengths of the three sides of the triangle.

Find the coordinates using the midpoint formula and length using the distance formula.

We have

Read and Learn More Savvas Learning Co Geometry Student Edition Solutions

We have,the​distance​between​(x₁,​y₁),​(x₂,​y₂) : \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)

To find AB, use A(5,1)and B(−3,3)

⇒  \(\sqrt{(-3-5)^2+(3-1)^2}\)

= 2\(\sqrt{17}\)

To find BC, use B(−3,3)and C(1,−7)

⇒  \(\sqrt{(1-(-3))^2+(-7-3)^2}\)

= 2\(\sqrt{29}\)

To find AC, use A(5,1)and C(1,−7)

⇒  \(\sqrt{(1-5)^2+(-7-1)^2}\)

= 4\(\sqrt{5}\)

The coordinates of the midpoint of the sides of AB, BC, AC are(1,2),(−1,−2), (3,−3)  respectively.  The length of the sides AB, BC, AC  are 2\(\sqrt{17}\), 2\(\sqrt{29}\),4 \(\sqrt{5}\) respectively.

Chapter 5 Relationships Within Triangles Exercise Savvas Geometry Answers Page 281  Exercise 3  Problem 3

Given: A(−1,2), B(9,2), C(−1,8)

To find – Find the coordinates of the midpoints of the sides and then find the lengths of the three sides of the triangle.

Find the coordinates using midpoint formula and length using distance formula.

We have

Midpoint​of​ (x₁,​y₁),​(x₂,​y₂):  \(\left(\frac{x_2+x_1}{2}, \frac{y_2+y_1}{2}\right)\)

To find AB, use A(−1,2) and B(9,2)

\(\left(\frac{9-1}{2}, \frac{2+2}{2}\right)\) = (4,2)

To find BC, use B(9,2)and C(−1,8)

\(\left(\frac{-1+9}{2}, \frac{8+2}{2}\right)\) = (4,5)

To find AC, use A(−1,2)and C(−1,8)

\(\left(\frac{-1-1}{2}, \frac{8+2}{2}\right)\) = (−1,5)

We have,the ​distance​between (x₁,​y₁),​(x₂,​y₂): \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
​
To find AB , use A(−1,2) and B(9,2)

⇒  \(\sqrt{(9-(-1))^2+(2-2)^2}\)

= 10

To find BC, use B(9,2) and C(−1,8)

⇒  \(\sqrt{(-1-9)^2+(8-2)^2}\)

= 2\(\sqrt{34}\)

To find AC, use A(−1,2)and C(−1,8)

⇒  \(\sqrt{(-1-(-1))^2+(8-2)^2}\)

= 6

The coordinates of the midpoint of the sides of AB, BC, AC are (4,2),(4,5),(−1,5) respectively. The length of the sides AB, BC, AC are 10, 2\(\sqrt{34}\), 6 respectively.

Relationships Within Triangles Exercises Solutions Chapter 5 Savvas Geometry Page 281  Exercise 4  Problem 4

Given: A(−2,−3), B(2,−3), C(0,3).

To find – Find the coordinates of the midpoints of the sides and then find the lengths of the three sides of the triangle.

Find the coordinates using the midpoint formula and length using the distance formula.

We have

Midpoint ​of (x₁,​y₁),​(x₂,​y₂): \(\left(\frac{x_2+x_1}{2}, \frac{y_2+y_1}{2}\right)\)

The midpoint of AB using A(−2,−3) and B(2,−3) is

\(\left(\frac{2-2}{2}, \frac{-3-3}{2}\right)\) = (0,−3)

The midpoint of BC using B(2,−3) and C(0,3) is

\(\left(\frac{0+2}{2}, \frac{3-3}{2}\right)\) = (1, 0)

The midpoint of AC using A(−2,−3) and C(0,3) is

\(\left(\frac{0-2}{2}, \frac{3-3}{2}\right)\) = (−1, 0)

We have the​ distance​between​ (x₁,​y₁),​(x₂,​y₂): \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
​
To find AB, use A(−2,−3) and B(2,−3) is

⇒  \(\sqrt{(2-(-2))^2+(-3-(-3))^2}\)

= 4

To find BC, use B(2,−3)and C(0,3)

⇒  \(\sqrt{(0-2)^2+(3-(-3))^2}\)

=  2\(\sqrt{10}\)

To find AC, use A(−2,−3) and C(0,3)

⇒  \(\sqrt{(0-(-2))^2+(3-(-3))^2}\)

=  2\(\sqrt{10}\)

The coordinates of midpoint of the sides of AB, BC, AC are (0,−3),(1,0),(−1,0) respectively. The length of the sides AB, BC, AC are 4, 2\(\sqrt{10}\),  2\(\sqrt{10}\) respectively.

Relationships Within Triangles Exercises Solutions Chapter 5 Savvas Geometry Page 281  Exercise 5  Problem 5

Given:  It is not too late.

To find –  The negation of a statement.

The negation of a statement is the opposite of the given mathematical statement.

It is too late is the negation of It is not too late.

The negation of the given statement is it is too late.

Chapter 5 Relationships Within Triangles Savvas Learning Co Geometry Explanation Page 281  Exercise 6 Problem 6

Given:  m∠R > 60

To find –  The negation of a statement.

The negation of a statement is the opposite of the given mathematical statement.

he measure of the angle R is not greater than 60.

The measure of the ∠R  is not greater than 60.

Solutions For Relationships Within Triangles Exercises In Savvas Geometry Chapter 5 Student Edition Page 281  Exercise 7  Problem 7

Given:  A(9,6), B(8,12)

To find – The slope of the line passing through the given points.

Slope between two points:

Slope = \(\frac{y_2-y_1}{x_2-x_1}\)

We have

⇒ (x₁,y₁) = (9,6)

⇒ (x₂,y₂) = (8,-12)

⇒ m = \(\frac{12-6}{8-9}\)

⇒ m = -6

The slope of the line passing through the given points is m =−6.

Relationships Within Triangles Exercises Savvas Learning Co Geometry Detailed Answers  Page 281  Exercise 8  Problem 8

Given: The distance between your home and your school is the length of the shortest path connecting them.

To find – How might you define the distance between a point and a line in geometry?

Distance is a numerical measurement of how far apart objects or points are.

The distance from a point to a line is the shortest distance from a given point to any point on an infinite straight line.

It is the perpendicular distance of the point to the line, the length of the line segment which joins the point to the nearest point on the line.

The distance from a point to a line is the shortest distance from a given point to any point on an infinite straight line. It is the perpendicular distance of the point to the line, the length of the line segment which joins the point to the nearest point on the line.

Savvas Learning Co Geometry Student Edition Chapter 5 Page 281  Exercise 9  Problem 9

Given:  A midpoint of a segment.

To find – What do you think a midsegment of a triangle is?

The midpoint is the middle point of a line segment.

It is equidistant from both endpoints, and it is the centroid both of the segment and of the endpoints.

A midsegment of a triangle is a segment that connects the midpoints of two sides of a triangle.

A midsegment of a triangle is a segment that connects the midpoints of two sides of a triangle.

Relationships Within Triangles Exercises Savvas Learning Co Geometry Detailed Answers Page 281  Exercise 10  Problem 10

Given: If two parties are happening at the same time, they are concurrent.

To find –  What would it mean for three lines to be concurrent?

Lines in a plane or higher-dimensional space are said to be concurrent if they intersect at a single point.

Three or more lines in a plane passing through the same point are concurrent lines.

When a third line also passes through the point of intersection made by the first two lines then these three lines are said to be concurrent lines.

The point of intersection of all these lines is called the ‘Point of Concurrency’.

Three or more lines in a plane passing through the same point are concurrent lines. When a third line also passes through the point of intersection made by the first two lines then these three lines are said to be concurrent lines. The point of intersection of all these lines is called the ‘Point of Concurrency’.